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% 12/11/2024 11AM

Also: what about products? If $c : \Nbb \to [k]$
then induce $c' : \Nbb \to [k]$ by $c'(n) =
c(2^n)$.

By the above for $c'$ you get $x_1, \ldots, x_n$
such that $c \left( \prod_{I \neq 0} 2^{x_i}
\right)$ is constant.

Question: Can we always fine $x_1, \ldots, x_n \in
\Nbb$ (when finitely coloured) such that the set
\[
  \left\{ \sum_{i \in I} x_i, \prod_{i \in I} x_i
  ~\forall I \subseteq [n], I \neq \emptyset \right\}
\]
is monochromatic?

This is very open \ldots even $n = 2$, i.e. $\{x,
y, x + y, xy\}$.

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If you insist on an infinite set
      $(x_i)_{i \in \Nbb}$, then you can find a
      bad colouring (Some new results on
      monochromatic sums and products over the
      rationals [Hindman, Ivan, Leader]).
    \item If we ask this question over $\Qbb$ --
      true (Alweiss, 2023+).
    \item It is also true that $\{x, x + y, xy\}$
      is \gls{pr} over $\Nbb$ (2023, Bowen and
      someone)
  \end{enumerate}
\end{remark*}

\begin{fcprop}[]
  \label{prop:2.8}
  % Proposition 2.8
  Assuming:
  - $A$ has the \gls{cp}
  Then:
  there exists $m, p, c$ such that any
  \gls{mpcset} contains a solution to $Ay = 0$.
\end{fcprop}

\begin{proof}
  Let $C_1, \ldots, C_n$ be the columns of $A$.
  Then there exists $B_1 \sqcup B_2 \sqcup \cdots
  \sqcup B_r$ a partition of $[n]$ such that
  \[
    \sum_{i \in B_k} C_i \in \Span_{\Qbb} (C_i : i
    \in B_1 \cup \cdots \cup B_{k - 1})
  .\]
  For all $k$, we have
  \[
    \sum_{i \in B_k} C_i = \sum_{i \in B_1 \cup
    \cdots \cup B_{k - 1}} q_{ik} C_i
  \]
  with $q_{ik} \in \Qbb$.

  For each $k$, let
  \[
    d_{ik} = \begin{cases}
      0 & \text{if $i \notin B_1 \cup \cdots \cup
      B_k$} \\
      1 & \text{if $i \in B_k$} \\
      -q_{ik} & \text{if $i \in B_1 \cup \cdots
      \cup B_{k - 1}$}
    \end{cases}
  \]
  Rewriting the above we get
  \[
    \sum_{i = 1}^{n} d_{ik} C_i = 0
  \]
  for all $k$. We will take $m = r$. Let $x_1,
  \ldots, x_r$ be some integers. Let $y_i = \sum_{k
  = 1}^{r} d_{ik} x_k$.

  Claim $(y_1, \ldots, y_n)^\top$ is a solution,
  i.e. $\sum_{i = 1}^{n} y_i C_i = 0$. Indeed:
  \begin{align*}
    \sum_{i = 1}^{n} y_i C_i
    &= \sum_{i - 1}^{n} \sum_{k = 1}^{r} d_{ik}
    x_k C_i \\
    &= \sum_{k = 1}^{r} \sum_{i = 1}^{n} d_{ik}
    x_k C_i \\
    &= \sum_{k = 1}^{r} x_k \left( \ub{\sum_{i =
    1}^{n} d_{ik} C_i}_{= 0} \right) \\
    &= 0
  \end{align*}
  Look at $y_i = \sum_{k = 1}^{n} d_{ik} x_k$.
  Have $d_{ik} \in \Qbb$. Let $c$ be the common
  denominator of all the $q_{ik}$s. Then
  \[
    cy_i = \sum_{k = 1}^{n} \ub{cd_{ik}}_{\in
    \Nbb} x_k
  \]
  Also have that $cy$ is a solution. Our $c$ (for
  the \mpc rpc) is indeed the common
  denominator of the $q_{ik}$, and $p = c \cdot
  \max|\text{numerators}|$.
\end{proof}

\begin{proof}[Proof of \nameref{rado}]
  Want to prove $A$ is \gls{pr} if and only if it
  has the \gls{cp}.

  If $A$ is \gls{pr}, then by \cref{prop:2.5}, it
  has the \gls{cp}.

  For the other direction, let $\tilde{c}$ be a finite
  colouring of $\Nbb$. Also, since $A$ has \gls{cp}
  there exists $m, p, c$ such that $Ax = 0$
  solutions in \emph{any} \gls{mpcset}. By
  \cref{thm:2.6} there exists a monochromatic
  \gls{mpcset} with respect to $\tilde{c}$. But
  this gives a monochromatic solution to $Ax = 0$.
\end{proof}

\begin{remark*}
  From the proof, we get that if $A$ is \gls{pr}
  for the ``$\bmod ~p$'' (right-most position in
  base $p$) colourings then in fact $A$ is
  \gls{pr} for any colouring. There is no direct
  proof of this (i.e. that does not go via the
  \nameref{rado} proof).
\end{remark*}

\begin{fcthm}[Consistency Theorem]
  \label{thm:2.9}
  % Theorem 2.9
  Assuming:
  - $A$ and $B$ two matrices that are \gls{pr}
  Then:
  \[
    \begin{pmatrix}
      A & 0 \\
      0 & B
    \end{pmatrix}
  \]
  is \gls{pr}.
\end{fcthm}

\begin{proof}
  Trivial check of \gls{cp}.
\end{proof}

This says that if you can solve $Ax = 0$
monochromatically and you can solve $Bx = 0$
monochromatically, then there exists $y_1, y_2$ of
the same colour such that $Ay_1 = 0$, $By_2 = 0$.

\begin{remark*}
  You can show this by hand (but much harder).
\end{remark*}

\begin{fcthm}[]
  \label{thm:2.10}
  % Theorem 2.10
  Assuming:
  - $\Nbb$ finitely coloured
  Then:
  a colour class contains solutions to all
  \gls{pr} equations.
\end{fcthm}

\begin{proof}
  $\Nbb = C_1 \sqcup C_2 \sqcup \cdots \sqcup
  C_k$. Assume for all $C_i$ that there exists
  $A_i$ that is \gls{pr}, but has no solution in
  $C_i$. Look at
  \[
    \begin{pmatrix}
      A_1 & 0 & \cdots & 0 \\
      0 & A_2 & \cdots & 0 \\
      \vdots & \vdots & \ddots & \vdots \\
      0 & 0 & \cdots & A_k
    \end{pmatrix}
  .\]
  This is \gls{pr} too, hence it has a
  monochromatic solution of colour say $C_t$. Then
  $A_t$ has a solution in $C_t$, contradiction.
\end{proof}

Rado's conjecture (1933)

\begin{definition*}
  $D \subseteq \Nbb$ is \gls{pr} if it contains
  solutions to all \gls{pr} equations.
\end{definition*}

Rado conjectured that if $D = D_1 \sqcup D_2
\sqcup  \cdots \sqcup D_k$, then one is also
\gls{pr}.

Proved in 1973 by Deuber -- introduced
\glspl{mpcset}. Showed that $D$ is \gls{pr} if and
only if it contains an \gls{mpcset} for all $m, p,
c$.

He then showed that given $m, p, c, k$, there
exists $n, q, d$ such that whenever an \mpc nqd is
$k$-coloured, there exists a monochromatic
\gls{mpcset} (this indeed solved the conjecture).